No.1397 (PR)

No.1397
Paul Rãican
(Румыния)

Оригинальные Retro & PG задачи
JF – R2019-20


Определения: (показать/спрятать)


No.1397 Paul Rãican
Romania

original – 16.04.2019
dedicated to quartzomaniacs

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 BPh2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 BPh7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 30                                       (13+16)
Duelist
Berolina Pawns


No.1397.1 Paul Rãican
Romania

version of No.1397 – 29.04.2019
dedicated to quartzomaniacs

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 BPh2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 BPh7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 30.5                                    (13+15)
Duelist
Berolina Pawns


4 комментария: No.1397 (PR)

  1. François Labelle пишет:

    Cooked by Jacobi “PG demolition mode” in 6h. The first 5.0 moves are the same as the intention.

    1.BPa4 BPb5 2.BPac4 BPbd5 3.BPe4 BPf5 4.BPg4 BPh5 5.BPb4 BPxb4 6.BPh4 BPa3 7.BPg5 BPce5 8.BPh6 BPf4 9.Sf3 BPxf3 10.Qd2 BPg2 11.Qf4 Rxh6 12.Qc7 Ra6 13.Qd7+ Kf7 14.Qc6 BPg5 15.Qd7+ Be7 16.Qc6 Bb4+ 17.BPc3 Bc5 18.BPd4 Bd6 19.BPe5 Bf8 20.BPf6+ Ke7 21.BPcb5 BPf4 22.Qc4 BPe3 23.Qb4+ BPc5 24.Qc4 BPb4 25.Qd4 BPc3 26.Qe5+ Re6 27.Qc7+ Ke8 28.Qb6 Kd7 29.Qd6+ Ke8 30.Qc6+ Ke7

    By the way, some moves of the text solution are missing disambiguation, for example 2.BPc4 BPd5. There is the same problem with the recent 1394 by Nicolas Dupont. A friendly reminder to composers using Jacobi to select the Output language “Py2Web format” and to send the solution in this format to Julia.

  2. Vlaicu Crisan пишет:

    For those who enjoy PG in Duellist with Berolina Pawns, please try to solve R261 from Problem Paradise 85/2019.

    Michel Caillaud, Vlaicu Crisan & Eric Huber
    rsb1k1s1/8/8/pppppppp/PPPPPPPP/8/3SB3/RSB1K1R1
    (15 + 13) PG 22.0 Berolina Pawn Duelist

    The Ber[o]lin[a] Wall.

  3. François Labelle пишет:

    1397.1 is cooked using PG demolition mode as well (in 8h this time).

    1.BPef3 BPh5 2.BPg4 BPhf5 3.BPce4 BPd5 4.BPc4 BPb5 5.BPb4 BPxb4 6.BPh4 BPa3 7.BPg5 BPce5 8.BPf6 BPf4 9.Sh3 BPg3 10.Sg5 BPxg2 11.Sf7 BPec5 12.Sxh8 BPd4 13.Sf7 BPe3 14.Sd6+ Bxd6 15.BPb5 Bb4+ 16.BPc3 Bc5 17.BPb4 Bd4 18.BPa5 Bb6 19.Qd4 Bc5 20.Qe5+ Be7 21.Qe6 BPc5 22.Qe5 BPb4 23.Qc3 Bf8 24.Qc6+ Ke7 25.Qd6+ Ke8 26.Qb6 Ke7 27.Qc6 Ra6 28.Qa8 Re6 29.Qc6 Rd6 30.Qc5 BPc3 31.Qc6

    The intention now shows 2.BPac4 BPbd5, but I see that someone removed the disambiguation suggested by Jacobi for 12… BPac5+ and 15… BPce5+, presumably because the alternative move is not check. The established convention is to disambiguate even in this case. One can verify by playing the orthodox game 1. d4 e5 2. dxe5 Ke7 3. Nd2 Ke6 4. a3 Kd5 5. Ndf3+ in any chess-playing program (N=knight).

  4. Paul Rãican пишет:

    Thank you, Francois! With Jacobi’s help, I found something else even shorter: 1.BPa4 BPb5 2.BPac4 BPbd5 3.BPe4 BPf5 4.BPg4 BPh5 5.BPb4 BPxb4 (till now as in intended solution) 6.BPh4 BPa3 7.BPg5 BPce5 8.BPf6 BPf4 9.Sf3 BPxf3 10.Qe2 BPg2 11.Qh6 BPec5 12.Qxh8 BPd4 13.Qh6 BPe3 14.Qg6+ Ke7 15.Qe8+ Kd6 16.Qb5 Ke7 17.Qc6 BPc5 18.Qb6 BPd4 19.Qa7+! Rxa7 20.BPab5 Rc7 21.BPa6 Rb7 22.BPc3 Rc7 23.BPb4 Rb7 24.BPa5 Rb6 25.BPb5 Rc6 26.BPb7 Rd6 27.BPa8=Q Re6 28.Qc6 Rd6 29.Qc5 BPc3 30.Qc6. Don’t need to say that Jacobi is fantastic.

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